SOLUTIONS TO ECE 6603 ASSIGNMENT NO. 6

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1 SOLUTIONS TO ECE 6603 ASSIGNMENT NO. 6 PROBLEM 6.. Consider a real-valued channel of the form r = a h + a h + n, which is a special case of the MIMO channel considered in class but with only two inputs. Assume a and a are independently and uniformly chosen from the binary alphabet {}, and assume that the noise components are real, independent and Gaussian with variance. This problem compares the jointly optimal (JML) detector, which jointly chooses a and a to maximize the conditional pdf f(r a, a ), to the individually optimal (IML) detector for the first input, which chooses a to maximize f(r a ). Let y = h T r, let y = h T r, and let = h T h. The JML detector chooses the pair (a, a ) T to maximize the conditional pdf f(r a, a ). It thus decides a = when max{f (r, ), f (r, )} > max{f (r, ), f (r, )} or equivalently, since the logarithm is monotonic: JML â = sign log---- maxfr fr. maxfr fr In contrast, the IML detector chooses its decision for a to maximize fra f(r a ) = + fra. fr a In other words, the IML detector chooses â according to the sign of log = -- : fr a = IML â = sign log- fr + fr. fr + fr Comparing, we see that the IML and JML detectors can be put into the same form: aˆ = sign log-- hfr fr, hfr fr where h JML (x, y )=max(x, y) for the JML detector, and h IML (x, y )=x + y for the IML detector. But f(r a, a ) = e r a h a h --- = --e r + h + h a e y + a y a a --- a y a y + a a = K e, where K is a constant, independent of a and a. Plugging this into the above yields:

2 aˆ y h Ke + y y Ke y + = sign log--- y h Ke + y + y Ke y y Ke y h e y e = sign log---, y -- y Ke h e + y e + where we used the fact that h(ax, Ay) = Ah(x, y) for both definitions of h(, ) and for any constant A. The above equation simplifies to: y aˆ = sign y h e y e +. log-- y h e + y e + (a) Show that the JML decision for a can be expressed as: JML â = sign y + -- g 0 (y ) -- g 0 (y + ), where we have introduced the nonlinearity g 0 (x) = x. Using the above with h JML (x, y )=max(x, y) yields: y â JML = sign y max e y e + log---- y max e + y e + y = sign y --- e log-- + y e = sign y + -- y -- y + = sign y + -- g 0 (y ) -- g 0 (y + ), where we have introduced the nonlinearity g 0 (x) = x.

3 (b) Show that the IML decision for a is also given by the above equation, but with g 0 (x) = x replaced by g (x) = log cosh(x/ ). This nonlinearity is sketched below: g (x) 0.5 = x For the special case of IML, where h IML (x, y) = x + y, the general result reduces to: â IML = sign y y e y e + + log---- y e + y e + + = sign y cosh y log y + cosh ---- = sign y + y logcosh ---- logcosh, y =sign y + -- g (y ) -- g (y + ), where g (x) = logcosh(x/ ).

4 (c) Show that g (x) x as 0. It follows that â IML and â JML usually agree at high SNR. The function g (x) is defined by: But e x / g (x) = e log x/ e + x/ = e log x / e + x /. goes to zero as 0, so that: g (x) = log e x / + e x / log() log e x = loge x / = log(e x ) = x. It follows that JML â and IML â usually agree at high SNR. (d) Find a value for y such that â IML 6= â JML, assuming y =, = and =. When y = = =, the JML and IML detectors reduce to: â JML = sign y, IML â = sign y + -- g (0) -- g () = sign y So any y satisfying -- IML JML g () < y < will result in â = but a =.

5 PROBLEM 6.. Consider the three-input memoryless channel H =[h,h,h 3 ], where h = [, 0,, 0] T, h = [,, 0, 0] T, and h 3 = [, 0,, ] T. Suppose the inputs are i.i.d. uniformly chosen from{}, and the channel adds real white-gaussian noise having the identity as an autocorrelation matrix. The QG decomposition of the matrix H = QG yields a lower-triangular G matrix whose first diagonal element G is positive and real. If y is the first component of the WMF output y = Q*r, then it may be expressed as: y = G a + n, where n is a Gaussian random variable with variance =. The probability that the stack sphere detector extends the node that does not belong to the transmitted path can be expressed as: P e = P( y G a > y + G a ). After substituting for y, this probability becomes: P( n > n + G a )= P(G + n a < 0) = P(n > G ) = Q( G ). The larger G is, the less likely it is that the sphere detector will extend the wrong branch from the root. (a) Assuming the stack sphere detector is used with the natural ordering, find a numerical value for the probability that the first node extended (besides the root node) is not a part of the actual transmitted path. In the natural ordering, the QG decompostion of H =[h, h, h 3 ] yields the following G matrix: G = Therefore, the probability of extending to the wrong half of the tree is: Q( G ) = Q( ) = %.

6 (b) Repeat part (a) based on the -3- ordering (which is the optimal BLAST ordering). In the -3- ordering, the QG decompostion of [ h, h 3, h ] yields the following G matrix: G = Therefore, the probability of extending to the wrong half of the tree with BLAST ordering is: Q( G ) = Q(.47) = %. (c) Based on a comparison of (a) and (b), argue qualitatively why the BLAST ordering reduces the complexity of the sphere detector tree search. The complexity of the sphere detector is directly related to how many nodes are visited by the search algorithm, and indirectly related to how quickly the cost threshold J is reduced. If the stack sphere detector extends to the correct node at the first stage, then it will search the correct side of the tree first, thereby establishing a low cost threshold J for its tentative decision. With this small cost threshold, searching the wrong side of the tree becomes less complex since many of its nodes will have costs greater than J, allowing them and their subbranches to be pruned. Comparing the answers from parts (a) and (b), we see that the stack sphere detector is half as likely to extend the wrong node at the first stage when the BLAST ordering is used instead of the natural ordering. Therefore, with the BLAST ordering, the stack sphere detector is half as likely to waste unnecessary computations searching the wrong side of the tree.

7 PROBLEM 6.3. Consider a -by- MIMO channel of the form r = a h + a h + n, where the components of the noise vector are i.i.d. CN(0, ), and where a and a are independently and uniformly chosen from the 4-QAM alphabet { j}. Suppose that the channel matrix is: H =[h,h ] =. 0 (a) ZF Find the weights w of the decentralized ZF detector for a, and ZF find the weights w of the decentralized ZF detector for a. These weights are the two rows of the channel inverse: H = 0 = 0 first row w ZF = [, 0] T (b) second row w ZF = [, ] T Find the weights w MF of the decentralized MF detector for a, and find the weights w MF of the decentralized MF detector for a. w MF = h = [, ] T w MF = h = [0, ] T (c) Find an equation for the SNR of the decentralized ZF detector for a. After the the ZF filter: y = ( w ZF ) T r = [, ]r = r r = (a + a + n ) (a + n ) E a = a + n n SNR = --- = =. E n n (d) Find an equation for the error probability of the decentralized ZF detector for a. After the the ZF filter: y = (w ZF ) T r = [, 0]r = r E a = a + n SNR = ---- =. E n -

8 For 4-QAM in AWGN with no ISI, the symbol-error probability is (see ECE660): P e = Q( ) Q SNR ( SNR ) Q( SNR ) = Q( - ). In the same conditions, and with Gray coding, the bit-error probability is (see ECE660): P b = Q( SNR ) = Q( - ). (e) Find an equation for the error probability of the decentralized MF detector for a. After the the MF filter: y = (w MF ) T r = [, ]r = r + r = (a + n ) + (a + a + n ) = a + a + n + n. Taking the real part yields: z = b + b + w + w, where z i = Re{y i }, b i = Re{a i } {±}, and w i = Re{n i } N(0, /). The decision about b will be incorrect when the sign of z disagrees with the sign of b. Equivalently, the decision will be incorrect when the product z b is negative: P b = P(z b < 0) = P((b + b + w + w )b < 0) = P( + b b + b w + b w < 0). Making the observation that the product b b {±} has exactly the same distribution as b alone, and similarly that the product b w i has exactly the same distribution as w i alone, we can simplify P b to: P b = P( + b + w + w < 0) = -- P( + + w + w < 0 b =+) + -- P( + w + w < 0 b = ) = -- P(3 + w + w < 0) + -- P( + w + w < 0) = -- P( (w + w ) > 3) + -- P( (w + w ) > ) = -- Q( 3 ) + -- Q( ).

9 (f) Find an equation for the single-user-bound (SUB) on error probability for both a and a. These can be derived from the SUB on the SNR, since there is no interference to consider in this case, and since the noise is AWGN; from ECE660 we learned that the relationship between bit-error probability and SNR in AWGN without interference for 4-QAM is: P b = Q( SNR ). The SUB on the SNR for the two symbols is: Therefore: SNR SUB = kh k / = -, SUB SNR = kh k / = -. SUB P b, = Q( - ), SUB P b, = Q( - ).

10 PROBLEM 6.4. A 5-input 0-output MIMO channel transmits an independent 6-QAM symbol from each of the 5 transmit antennas. This question is about the structure of the detection tree. (a) How many layers are in the detection tree? One for each transmitter n = 5 layers. (b) How many branches emanate from each non-leaf node in the detection tree? One for each candidate symbol A = 6 branches. (c) How many leaf nodes are in the detection tree? One for each candidate symbol vector A n = 6 5 = 0 = > million. (d) How many total nodes (including the root node and leaf nodes and all in between) are in the detection tree? Including the single root node at layer 0, the total number of nodes in the tree is: P n N = =0 A P 5 = =0 6 = = 848 total nodes. (e) How many branches in all are in the detection tree? There is precisely one branch leading into each nonroot node. Subtracting one from the answer in part (d), to exclude the root node, we get: 8480 total branches.